Try to justify those parameters.sol59 said:I've tried using the Poisson distribution with k=2 t=40/60=2/3 and lambda=4/3
haruspex said:Try to justify those parameters.
In the equation you quote, how would you describe the meanings of k and λ?
Yes, λ here is 2 per hour.sol59 said:I think k is number of events that can happen in selected time unit...two student arrivals in this case and λ is average number of events that can happen in an hour...is it 2 as well? I'm not sure.
haruspex said:Yes, λ here is 2 per hour.
But the equation you quote is not suited to the task, directly. It gives you the probability of a specific number of arrivals in a specific period of time. You want the probability that the time between a particular pair of consecutive arrivals is in some range.
You might as well suppose that the first of the two arrivals has just occurred. How can you rephrase the question a bit more simply?
haruspex said:Yes, λ here is 2 per hour.
But the equation you quote is not suited to the task, directly. It gives you the probability of a specific number of arrivals in a specific period of time. You want the probability that the time between a particular pair of consecutive arrivals is in some range.
You might as well suppose that the first of the two arrivals has just occurred. How can you rephrase the question a bit more simply?
No, you need to do some more thinking before trying to plug numbers into an equation.sol59 said:2*e(-2*t) where t=2/3?
haruspex said:No, you need to do some more thinking before trying to plug numbers into an equation.
Suppose that one student has just arrived. Rephrase the question in terms of a single arrival.
No, we can start the clock when the first student arrives. What are the constraints on the arrival time of the next student?sol59 said:Well I need to find out if the time of waiting for arrival of the second student is in the interval <= 50 minutes. And the arrival of the first student has to be in the interval >= 10 minutes.
haruspex said:No, we can start the clock when the first student arrives. What are the constraints on the arrival time of the next student?
No, that is not what the question says. Read it again:sol59 said:The second student has to come 40 minutes later at most after the first student.
Edit: maybe the word interval is confusing you. Here it just means a range of values: the time between two consecutive arrivals is in the range from 10 minutes to 50 minutes.sol59 said:the time between two consecutive arrivals is in the interval <10 minutes; 50 minutes>.
haruspex said:No, that is not what the question says. Read it again:
Edit: maybe the word interval is confusing you. Here it just means a range of values: the time between two consecutive arrivals is in the range from 10 minutes to 50 minutes.
And the earliest?sol59 said:The second student has to come in the fiftieth minute at the latest?
haruspex said:And the earliest?
No, what is the earliest, after the first student, that the second student can arrive (to fit in the given window)?sol59 said:The fiftieth minute minus the time when the first student came?
haruspex said:No, what is the earliest, after the first student, that the second student can arrive (to fit in the given window)?
Let's make it more concrete. Say the first student arrives at 9am. Between what two times is the second student to arrive for the gap between them to be from 10 to 50 minutes?
Right.sol59 said:9:10-9:50?
haruspex said:Right.
So how many students arrive between 9 and 9:10?
Right. Using your equation, what is the probability of that?sol59 said:0
No. We might come back to that later.sol59 said:and between 9:50-10 too)
haruspex said:Right. Using your equation, what is the probability of that?
No. We might come back to that later.
How many students arrive between 9:10 and 9:50?
We are getting there. But why exactly one student between 9:10 and 9:50? What goes wrong if two arrive?sol59 said:9:00-9:10 P(Y=0)=((2*1/6)0*(e-1/3))/0!=0.716
9:10-9:50 1 student P(Y=1)=((2*2/3)1*(e-4/3)/1!=0.351
haruspex said:We are getting there. But why exactly one student between 9:10 and 9:50? What goes wrong if two arrive?
For the moment, just try to answer my question. If no student arrives between 9 and 9:10, but two arrive between 9:10 and 9:50, does that meet the requirement that the time from the 9am arrival to the next arrival lies between 10 minutes and 50 minutes?sol59 said:I don't know:( P(Y=2)=0.234 but I don't understand what to do with it
haruspex said:For the moment, just try to answer my question. If no student arrives between 9 and 9:10, but two arrive between 9:10 and 9:50, does that meet the requirement that the time from the 9am arrival to the next arrival lies between 10 minutes and 50 minutes?
What if 3 arrive between 9:10 and 9:50? Four? ...
Right. So is there any limit on the number that can arrive between 9:10 and 9:50?sol59 said:I think it meets that requirement
That's just an average rate. Poisson processes have no memory. In each short period of time dt, the probability of an arrival is λdt. If one does arrive, the probability is still λdt for the next period dt. In principle, a million could arrive in a single minute (ok, this is an idealised view), but the probability would be vanishingly small.sol59 said:but it is also said that 2 students per hour come into the class.
haruspex said:Right. So is there any limit on the number that can arrive between 9:10 and 9:50?
That's just an average rate. Poisson processes have no memory. In each short period of time dt, the probability of an arrival is λdt. If one does arrive, the probability is still λdt for the next period dt. In principle, a million could arrive in a single minute (ok, this is an idealised view), but the probability would be vanishingly small.
What number of arrivals between 9:10 and 9:50 does not meet the condition?sol59 said:That's interesting...so there is an unlimited number of students that can arrive in that interval. But how to express it mathematically?
haruspex said:What number of arrivals between 9:10 and 9:50 does not meet the condition?
Right.sol59 said:0?
haruspex said:Right.
So we have boiled it down to two requirements:
No arrivals in (9:00, 9:10), and
Not (no arrivals in (9:10, 9:50))
Can you figure out the probability of the second of those, and combine it with the probability you already found for the first?
Right, so what is the probability of at least 1 arrival in (9:10, 9:50) ?sol59 said:The probability of no arrivals in (9:10, 9:50) is P(Y=0)=e-4/3=0.264
haruspex said:Right, so what is the probability of at least 1 arrival in (9:10, 9:50) ?
Good. So put it all together. What is the probability of no arrivals in (9:00, 9:10) and at least one in (9:10, 9:50)?sol59 said:1-0.264=0.736
haruspex said:Good. So put it all together. What is the probability of no arrivals in (9:00, 9:10) and at least one in (9:10, 9:50)?
You got there!sol59 said:0,736*0.716=0.527
haruspex said:You got there!
